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# Show that if available carpentry hours remain between 60 and 100, the current basis remains optimal. If between 60 and 100 carpentry hours are available, would Giapetto still produce 20 soldiers and 60 trains?

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We are to show that if available carpentry hours remain between $60$ and $100,$ the current basis remains optimal. Also we are to check if Giapetto would still produce $20$ soldiers and $60$ trains. Ofcourse, maximizing the objective function is $\max z=3x_1+2x_2,$ where $x_1$ and $x_2$ are numbers (nonnegative) of soldiers and trains produced per week, respectively. It is subjected to previous constraints:

$2x_1+x_2\leq100,x_1+x_2\leq80,x_1\leq40.$

Let us $b_2$ be the number of carpentry hours available. Obviously the change in $b_2$ shifts the carpentry constraint parallel to its current position. However, as long as both constraints intersect in the feasible region the current basis will remain optimal. From the figure preented below we see that if $b_2<60$ intersection of finishing and carepntry constraint is not in the feasible region, thus the current basis is not optimal. We see that for $b_2=60$ the point $(40,20)$ is intersection of all $3$ constraints (in feasible region ofcourse). If however $b_2<60$ the demand constraint would not be satisfied. If $b_2>100$ then finishing and carpentry constraint would never intersect, thus the current basis would not remain optimal. So the current basis will remain optimal if $60\leq b_2\leq100.$ However, the optimal solution will change. Let us $b_2=80+\Delta.$ The current basis remains optimal if $-20\leq\Delta\leq20.$ Thus it has to be:

$2x_1+x_2=100,x_1+x_2=80+\Delta,$

since we want to find a point where the constraints are binding. Subtracting the second from the first equation yields

$x_1=20-\Delta,x_2=60+2\Delta.$

Simply, increasing available carpentry hours decreases at the same rate the optimal number of soldiers produced. Similarly the number of trains increases two times the number of additional carpentry hours. Therefore, we are done.

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